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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 10: Solutions Exponential and Gaussian Random Variables; Poisson Processes 1. [Reliability function of a triplemodularredundancy (TMR) system] (a) It is easy to see that { Y > T } = [ { X 1 > T } ∩ { X 2 > T } ∩ { X 3 > T } ] ∪ [ { X 1 > T } ∩ { X 2 > T } ∩ { X 3 ≤ T } ] ∪ [ { X 1 > T } ∩ { X 2 ≤ T } ∩ { X 3 > T } ] ∪ [ { X 1 ≤ T } ∩ { X 2 > T } ∩ { X 3 > T } ] Since the four events (on the four lines above) are mutually exclusive, we have that R Y ( t ) = P { Y > t } = [ R ( t )] 3 + 3[ R ( t )] 2 [1 R ( t )] = 3[ R ( t )] 2 2[ R ( t )] 3 . (b) Since R ( t ) = exp( λt ), we get that R Y ( t ) = 3exp( 2 λt ) 2exp( 3 λt ). Hence, E [ Y ] = Z ∞ R Y ( t ) dt = Z ∞ 3exp( 2 λt ) 2exp( 3 λt ) dt = 3 2 λ 2 3 λ = 5 6 λ < 1 λ = E [ X i ]. Next, let x = exp( λt ). Then, R Y ( t ) = 3 x 2 2 x 3 = 1 2 , which has only one solution x = 1 / 2 in the range 0 ≤ x ≤ 1. Thus, the median value of Y is the solution to x = exp( λt ) = 1 / 2, giving t = λ 1 ln2....
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 Normal Distribution, Poisson Distribution, Ry, −3λt).

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